Answer
Standard form: $f(x)=(x-1)^2-1$
Vertex: $(1,-1)$
Axis of symmetry: $x=1$
The x-intercepts: $(2,0)$ and $(0,0)$
Work Step by Step
$f(x)=x^2-2x~~$ ($a=1,~b=-2,~c=0$)
$f(x)=x^2-2x+1-1$
$f(x)=x^2-2(1)x-1^2-1$
$f(x)=(x-1)^2-1$
$-\frac{b}{2a}=-\frac{-2}{2(1)}=1$
$f(1)=(1-1)^2-1=-1$
Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(1,-1)$
So, the axis of symmetry is $x=-\frac{b}{2a}=1$.
The x-intercepts:
$f(x)=0$
$(x-1)^2-1=0$
$(x-1)^2=1$
$x-1=±1$
$x=2$ or $x=0$
So, the x-intercept points are: $(2,0)$ and $(0,0)$
